a) From symmetry of the problem we have
and from mirror symmetry we have
Hence
Hence
b) Consider a cylinder with its axis oriented along the z axis
and completely lying in the
half of the space. Let the base of the cylinder be at
and the top at
, the flux of electric field through the surface of the cylinder
is
Because of the form of electric field mentioned in equation (1)
the dot product between the area vector and electric field is
non-zero only at the base and top of the cylinder. Integral over
signifies integral over the area of the top or bottom of the
cylinder. Since this cylinder encloses no charge, integral form of
Gauss law gives us
We have proved equation (6) for arbitrary positive numbers
, for arbitrary negative numbers similar construction and similar
arguments can be used.
c) Using results of part (b) we get
and
Hence
HW.37. Redo Hw. 27 using the sump condition (0) and Gauss's law for Ē. As we...